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Algebra 2 AP Calculus AB Contact Info Pre-Calculus. Suggested Activities Week of March 23rd pg. 254-257: 1-11, 43, 45-48, 50-51 A33: Lesson 4-7: 1-6. Alright so I am taking 8 week classes instead of 16 because I am way behind but this summer I'll be able to finally take calc 1 then I can start my engineering courses. The only problem is instead of a 12 week class it's 8 week and I'm terrified of it. Taking two precalculus classes now, wrapping up algebra then trig.

Test your readiness for the AP Calculus exam with this quiz!

Answer 1

**A****: **If the graph of a function has a sharp point, the function is not differentiable at that point. This is because the slopes directly to the left and right of the point do not approach the same value.

An absolute value function has a sharp point at its vertex. The graph off(x)= x + 4 is a horizontal translation (to the left 4 units) of the standard absolute value function, y = x , which has vertex (0, 0). Thus, the vertex of fis (-4, 0), which means the function is not differentiable at x =-4. Choice (A) is correct.

Answer 2

**D:** The only discontinuities that are removable are holes and holes with a point above or below—this function has neither. The function has 2 jump discontinuities (gaps), at x=-2 and x= 0, but neither of these is removable.

Answer 3

**D:** Examine the values in the table: f(x) increases as x gets larger, which indicates that f′(x), the slope of the function, is positive. This means f′(x)> 0, so eliminate (A) and (B).

To choose between (C) and (D), take a closer look at the slopes. The slope between the first pair of points is 3, and the slope between the second pair of points is also 3, so f′(x) is constant. This means f′′(x), which is the derivative of f′(x), must be 0. Choice (D) is correct

Answer 4

**C:** To use a local linear approximation, you need to find the equation of the tangent line. You’ve been given all the information you need in the question stem; you just need to piece it all together. The point on the function is given by f(-1)= 5, which translates to the point (-1, 5). The slope of the tangent line at x=-1 is given by f'(-1)= 2, so the slope is 2. Now the point-slope form of the tangent line is:

y-5 = 2(x-(-1))

y = 5+2(x+1)

Substituting x=-0.9 into the equation of the tangent line yields:

5+2(-0.9+1) = 5+2(0.1) = 5.2

That’s (C).

For more practice questions, check out our AP Calculus Prep Plus book.

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